Given a bounded domain $\Omega\subset \mathbb{R}^n$ and a smooth function $f$ with bounded derivatives on $\Omega$, is it possible to extend $f$ to $\tilde{f} : \mathbb{R}^n \to \mathbb{R}$ such that it is smooth and compactly supported with given closed set $\Omega ' \supset \Omega$ ?
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chandu1729
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1If your function has uniformly bounded derivatives to all orders, it is in fact real analytic, see http://calculus.subwiki.org/wiki/Uniformly_bounded_derivatives_implies_globally_analytic – Willie Wong May 03 '13 at 09:08
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After the question has been edited, the answer is yes, if the boundary of $\Omega$ is sufficiently smooth and there is an open set $G$ with $\overline{\Omega}\subset G\subset\Omega´$. Note at first, that $f$ can be continuously extended to $\overline{\Omega}$ by a classical topological result for uniformly continuous functions. And as the same is true for all partial derivatives, this extension is infinitely continuously differentiable.
The smooth extension to all of $\mathbb{R}^n$ is basically Whitney's extension theorem, see e.g. this question. And multiplication with a suitable function gives the compact support.
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No, consider $f: (0, 1) \to \mathbb{R}$, $f(x) = \frac{1}{x}$. You cannot extend $f$ over $0$. Generally, the problem of extending functions defined on open sets is hopeless.
xyzzyz
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Thanks. But, what if we impose additional condition that $f$ is is $\mathit{C}^\infty_b$ ? – chandu1729 Apr 28 '13 at 19:50
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You mean, $f$ is smooth and bounded? Still hopeless, consider $f: (0, 1) \to (0, 1)$, $f(x) = \sin(\frac{1}{x})$ – xyzzyz Apr 28 '13 at 19:51
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